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(3p² + q²) (2p² – 3q²)

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c – d) = a × (c – d) + b × (c – d) = (a × c – a × d) + (b × c – b × d)

= ac – ad + bc – bd

Let,

a= 3p², b= q², c= 2p², d= 3q²

Now,

= 3p²× (2p² – 3q²) + q² × (2p² – 3q²)

= [(3p²× 2p²) + (3p²× -3q²)] + [(q² × 2p²) + (q² × -3q²)]

= [6p⁴ – 9p²q² + 2q²p² – 3q⁴)]

= [6p⁴ – 7p²q² – 3q⁴]

Correct option is  C) 8400

Correct option is (C) 8400

\(92^2-8^2=(92-8)(92+8)\)

= 84 \(\times\) 100 = 8400.

Correct option is   A) \(\frac{PTR}{100}\)

Using the identity 

(a + b)² = a² + 2ab + b² 

(x² +5)² = (x²)² + 2.x².5 + 52 

= x⁴ + 10x² + 25

Using the identity 

(a + b)² = a² + 2ab + b² (2p + 3q)² 

= (2p)² + 2.2p.3q + (3q)² 

= 4p² + 12pq + 9q²

Using the identity 

(a + b)² = a² +2ab + b² 

(3x + 2y)² =(3x)² + 2.3x.2y + (2y)² 

= 9x² +12xy + 4y²

i) (x + a) (x + b) = x² + (a + b)x + ab

x = 2, a = 1, b = 3 then

⇒ (2 + 1) (2 + 3) = 2² + x(1 + 3)² + 1 × 3

⇒ 3 × 5 = 4 + 4x² + 3

⇒ 15 = 4 + 8 + 3 

⇒ 15 = 15

∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then

⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2

⇒ 1 × 2 = 0 + 0 + 2

⇒ 2 = 2

∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.

Using the identity 

(a + b)² = a² + 2ab + b²(a + 6)² 

= a² +2.a.6 + 6² = a² +12a + 36

(x + a) (x + b) (x + c) 

Consider (x + a)(x + b) 

= x² + (a + b)x + ab 

= x² + ax + bx + ab (x + a)(x + b)(x + c) 

= (x² + ax + bx + ab)(x + c) 

= x(x² + ax + bx+ ab) + c (x² + ax + bx + ab) 

= x³ + ax² + bx² + abx + cx² + cax +cbx + abc 

= x³ + ax² + bx²+ cx² + abx + cax +cbx + abc [by rearranging] 

= x³ + x²(a + b + c) + x(ab + ca +bc) + abc 

∴ (x + a)(x + b)(x + c) = x³ +x² (a + b + c) + x(ab + bc + ca) + abc

103 × 96 = (100 + 3)(100 - 4)

This is in the form. 

(x + a)(x + b) = x² + (a + b)x + ab 

Here x = 100, a = 3, b = -4 

(100 + 3)(100-4) = 100² + (3 – 4)100 + 3 × -4 

= 10000 + (- 1)100 – 12 

= 10000 – 100 – 12 

103 × 96 = 9888

i. (a + 3) (a + 5) 

This is in the form. 

(x + a)(x + b) = x² + (a + b)x + ab 

Here x = a, a = 3, b = 5 

(a + 3)(a+ 5) = a² +(a + 5)a + 3.5 

= a² + 8a + 15 

(ii) (3t + 1)(3t + 4)

This is in the form. (x + a)(x + b) 

= x² +(a + b)x + ab 

Here x =3t,a = 1, b = 4 

(3t – 1)(3t + 4) = (3t)²+(1+ 4)3t + 1.4 

= 9t² +5(3t) + 4 = 9t² +15t + 4 

(iii) (a -8)(a + 2)

This is in the form (x + a)(x + b) = x² +(a + b)x + ab 

Here x = a, a = -8, b = 2 

(a – 8)(a + 2) = a² + (-8 + 2)a + (-8)(2) 

= a² + (-6)a – 16 = a² – 6a – 16 

(iv) (a – 6)(a – 2)

This is in the form (x + a)(x + b) 

= x² +(a + b)x + ab 

Here x = a, a = -6,b = -2 

(a – 6)(a – 2) = a² + (- 6 – 2) a + (-6)(-2) 

=a² + (- 8)a + 12 

= a² – 8a + 12

53 × 55 = (50 + 3) (50 + 5) 

This is in the form. 

(x + a)(x + b) = x² + (a + b)x + ab 

Here x = 50, a = 3, b = 5 

(50 + 3)(50 + 5) = 50² + (3 + 5)50 + 3 × 5 

= 2500 + (8)50 +15 

= 2500 + 400 + 15 

∴ 53 × 55 = 2915

34 × 36 = (30 + 4) (30 + 6) 

This is in the form. 

(x + a)(x + b) = x²+ (a + b)x + ab 

Here x =30, a = 4, b = 6 

(30 + 4)(30 + 6) = 30² + (4 + 6)30 + 4 × 6 

= 900 + 300 + 24 

34 × 36 = 1224

102 × 106 = (100 + 2) (100 + 6) 

This is in the form. (x + a)(x + b) 

= x² +(a + b)x + ab 

Here x = 100, a = 2, b = 6 

(100 + 2)(100 + 6) = 100² +(2 + 6)100 + 2 × 6 

= 10000 + (8)100 + 12 

= 10000 + 800 + 12 

∴102 × 106 = 10812

Correct option is  B) 1

Correct option is (B) 1

a = 1, b = 2

Then \((a-b)^2\) \(=(1-2)^2=(-1)^2=1.\)

Correct option is  B) 8

Correct option is (B) 8

The value of 3x + 5 at x = 1

\(=3\times1+5=3+5=8.\)

Correct option is  B) 9964

Correct option is (B) 9964

\(106\times94=(100+6)(100-6)\)

\(=100^2-6^2\) = 10000 - 36 = 9964.

Correct option is  B) – 6x² 

Correct option is (B) –6x²

\(–3x^2+3x^2-5x^2-x^2\) \(=-5x^2-x^2\) \(=-6x^2.\)

Correct option is  D) None

A) 3x² , 2x² , – x² 

Given (x² – xy + y²) × (x + y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x² – xy + y²) × (x + y)

⇒ x (x² – xy + y²) + y (x² + xy + y²)

⇒ x³ – x²y – y²x + x²y + y²x + y³

⇒ (x³+ y³)

Given (x² + xy + y²) × (x – y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x² + xy + y²) × (x – y)

⇒ x (x² + xy + y²) + y (x² + xy + y²)

⇒ x³ + x²y + y²x – x²y – y²x + y³

⇒ (x³– y³)

 Given 24x²y³ by 3xy

⇒24x²y³ / (3xy)

On dividing monomial by a monomial we have divide same variables of each

Expressions.

On simplifying we get,

⇒8x²y

Given 7x,-3x, 5x, -x, -3x

To add the given expression we have arrange them column wise is given below:

7x

-3x

5x

-x

-3x

Given 8ab, -5ab, 3ab, -ab

To add the given expression we have arrange them column wise is given below:

8 ab

-5 ab

3 ab

-ab

0.5 × \(\frac{1}{3}\)× 24 × x × x × y² × y × x² × z⁴ × z

= \(\frac{12}{3}\)× x⁴ × y³ × z⁵

= 4x⁴ × y³ × z⁵

= 4x⁴y³z⁵

Given -2abc from -8abc

According to the rules of subtraction of algebraic equations, we have negative sign will becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows and columns we get

-8abc

-2abc

+

The given expression is

(ab - c)² + 2abc

= (ab)²-2(ab)(c) + (c)² + 2abc

= a²b²-2abc + c² + 2abc

= a²b² + c²

Therefore, (ab - c)² + 2abc = a²b² + c²

Given -16p from -11p

According to the rules of subtraction of algebraic equations, we have negative sign will becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows and columns we get

-16p

-11p

+

(a) The distance between Earth and Moon is 384,000,000 m.

= 3.84 × 10⁸ m

(b) Speed of light in vacuum is 300,000,000 m/s

= 3 × 10⁸ m/s

(c) Diameter of the Earth is 1,27,56,000 m.

1.2756 × 10⁷ m

(d) Diameter of the Sun is
1,400,0, 000m.

1.4 × 10⁹m

(e) Number of average stars in a galaxy = 100,000,000m.

= 1 × 10¹¹ m

(f) The universe is estimated to be about 12,000,000,000 years old.

= 1.2 × 10¹⁰ years old

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000, 000, 000 m.

= 3 × 10²⁰ m

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

= 6.023 × 10²² molecules

(i ) The earth has 1,353,000,000 cubic km of sea water.

= 1.353 × 10⁹ cubic km

(j) The population of India was about 1,027,000000 in March, 2001.

= 1.027 × 10⁹

\(\frac{4}{3}\)× \(\frac{1}{4}\) × 16 × p × p² × p² × q² × q² × r × r²

= \(\frac{-16}{3}\)× p⁵ × q⁴ × r³

= \(\frac{-16}{3}\)p⁵q⁴r³

(i) 5,00,00,000

5 × 10⁷

(ii) 70,00,000

7 × 10⁶

(iii) 3,18,65,00,000

3.1865 × 10⁹

(iv) 3,90,878

3.90878 × 10⁵

(v) 39087.8

3.90878 × 10³

(vi) 3908.78

3.90878 × 10³

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰

= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1

= 80,000 + 6000 + 0 + 40 + 5

= 86,045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰

= 400000 + 5000 + 300 + 2

= 4,05,302

(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰

= 40000 + 500 + 5

= 130,7051

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹

= 900000 + 200 + 30

= 9,00,230

(x² – y²) from (2x² – 3y² + 6xy)

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (2x² – 3y² + 6xy) – (x² – y²)

Change the sign of each term of the expression to be subtracted and then add.

= 2x² – 3y² + 6xy – x² + y²

= (2x² – x²) + (– 3y² + y²) + 6xy

= (2 – 1)x² + (– 3 + 1)y² + 6xy

= 1x² + (– 2y²) + 6xy

= 1x² – 2y² + 6xy

x² from -3x²

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

Then,

= (-3 – 1)x²

= -4x²

(a² + b² – 2ab) from (a² + b² + 2ab)

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (a² + b² + 2ab) – (a² + b² – 2ab)

Change the sign of each term of the expression to be subtracted and then add.

= a² + b² + 2ab – a² – b² + 2ab

= (1 -1)a² + (1 – 1)b² + (2 + 2)ab

= (0)a² + (0)b² + (4)ab

= 4ab

(i) 279404

= 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1

= 2× 10⁵ + 7× 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

(ii) 3006194

= 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1

= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰

(iii) 2806196

= 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10¹ + 6 × 10⁰

= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰

(iv) 120719

= 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1

= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰

(v) 20068

= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1

= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Given 7 + x − x², take away 9 + x + 3x² + 7x³

= (7 + x – x²) – (9 + x + 3x² + 7x³)

= 7 + x – x² – 9 – x – 3x² – 7x³

= – 7x³– x² – 3x² + 7 – 9

= – 7x³ – 4x² – 2

(x – y) from (4y – 5x)

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficient of the two like terms.

We have,

= (4y – 5x) – (x – y)

Change the sign of each term of the expression to be subtracted and then add.

= 4y – 5x – x + y

= (-5x – x) + (4y – y)

= -5x + 3y)

= 3y – 5x

(a) a (a² + a + 1) + 5 = a³ + a² + a + 5

(i) For a = 0, a³ + a² + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a³ + a² + a + 5 = (1)³ + (1)² + 1 + 5 = 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a³ + a² + a + 5 = (−1)³ + (−1)² + (−1) + 5 = − 1 + 1 − 1 + 5 = 4

Give. (x⁴+(1/x⁴) × ( x + (1/x))

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x⁴+(1/x⁴) × ( x + (1/x))

⇒ x⁴( x + (1/x)) + (1/x⁴) ( x + (1/x))

⇒ x⁵+x³+(1/x³)+(1/x⁵)

Given (x² -3x + 7) × (2x + 3)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x² -3x + 7) × (2x + 3)

⇒ 2x (x² -3x + 7) + 3 (x² -3x + 7)

⇒ 2x³ – 6x² + 14x + 3x² + 9x +21

⇒ 2x³ – 3x² +5x +21

Given (3x² + 5x – 9) × (3x – 5)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3x² + 5x – 9) × (3x – 5)

⇒ 3x (3x² + 5x – 9) – 5 (3x² + 5x – 9)

⇒ 9x³ + 15x² – 27x – 15x² – 25x + 45

⇒ 9x³ – 52x + 45

Number of terms in a monomial is 1.

Expression with one term is called a ‘Monomial’.

The no. of terms in the product of a monomial and a binomial are two (2).

Given that, x + \(\frac{1}{x}\) = 9

Squaring both sides, we get

(x + \(\frac{1}{x}\))² = 9²

x² + \(\frac{1}{x\times x}\) + 2 = 81

x² + \(\frac{1}{x\times x}\) = 79

Again,

Squaring both sides, we get

(x² + \(\frac{1}{x\times x}\) )² = 79²

x⁴ + \(\frac{1}{x\times x\times x\times x}\) + 2 = 6241

x⁴ + \(\frac{1}{x\times x\times x\times x}\) = 6239

33 × 27 = (30 + 3) (30 – 3) 

identity (a + b)(a – b) = a² – b² 

Here a = 30,b = 3 

(30 + 3)(30 – 3) = 30² – 3² = 900 – 9 

33 × 27 = 891

(2m – l) (2l – m) 

= 2m(2l – m) – l(2l – m)

= 2m × 2l – 2m × m – l × 2l + l × m

= 4lm – 2m² – 2l² + lm

= 5lm – 2m² – 2l²

(2x – y)(2x + y) (4x² + y²) 

Identity (a + b)(a – b) = a² – b² 

[(2x)² – y²](4x² + y²) 

= (4x² – y²)(4x² + y²) 

= 16x⁴ – y⁴

a² – a – 6 

= (- 3)² – (- 3) – 6 (when a = – 3) 

= (- 3 × – 3) + 3 – 6 

(Error is – (- 3) = – 3) 

= 9 + 3 – 6 

= 12 – 6 = 6